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Chapter 3

Chapter 3. Elementary Number Theory and Methods of Proof. 3.6. Indirect Argument. Reductio Ad Absurdum. Argument by contradiction Illustration in proof of innocence Suppose I did commit the crime. Then at the time of the crime, I would have had to be at the scene of the crime.

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Chapter 3

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  1. Chapter 3 Elementary Number Theory and Methods of Proof

  2. 3.6 Indirect Argument

  3. Reductio Ad Absurdum • Argument by contradiction • Illustration in proof of innocence • Suppose I did commit the crime. Then at the time of the crime, I would have had to be at the scene of the crime. • In fact, at the time of the crime, I was meeting with 20 people far from the crime scene, as they will testify. • This contradicts the assumption that I committed the crime, since it is impossible to be in two places at one time. Hence, that assumption is false.

  4. Proofs • Direct Proof • start with hypothesis of a statement and make one deduction after another until the conclusion is reached. • Indirect Proof (argument by contradiction) • show that a given statement is not true leads to the contradiction.

  5. Example • Use proof by contradiction to show that there is no greatest integer. • Starting point: Suppose not. Suppose that there is a greatest integer, N. N≥n for all integers. • To Show: This supposition leads logically to a contradiction. • Proof: • Suppose not. Suppose that there is a greatest integer N. • N ≥ nfor every integer n. Let M = N + 1. M is an integer under the addition property of integers. • Thus M > N. N is not the greatest integer; therefore a contradiction. • Theorem 3.6.1

  6. Even and Odd Integer • Theorem 3.6.2 • There is no integer that is both even and odd • Proof: • Suppose not. That is, suppose there is an integer n that is both even and odd. • n = 2a (even) and n = 2b + 1 (odd) • 2a = 2b + 1, 2a – 2b = 1, • 2(a – b) = 1 • (a – b) = ½ • Since a and b are integers then a – b should result in an integer. Thus a-b is integer and a-b is not, contradiction.

  7. Sum of Rational and Irrational • Theorem 3.6.3 • The sum of any rational number and any irrational number is irrational • Proof: • Suppose not. Suppose there is a rational number r and irrational number s such that r + s is rational. • r = a/b (definition rational) • r + s = c/d, for integers a,b,c,d with b≠0 and d≠0 • a/b + s = c/d, s = c/d – a/b • s = (bc – ad)/bd (integer) • hence, s is quotient of integers and therefore, rational • If s is rational it contradicts supposition that it is irrational

  8. Argument by Contrapositive • Argument by contrapositive is based on the equivalence of the statement and contrapositive. • If the contrapositive is true then the statement is true.

  9. Proof by Contrapositive • Method of Proof • Express the statement to be provided in the form: ∀x in D, if P(x) then Q(x) • Rewrite the statement in the contrapositive form: ∀x in D, if Q(x) is false then P(x) is false • Prove the contrapositive by a direct proof. • Suppose x is a (particular but arbitrarily chosen) element of D such that Q(x) is false. • Show that P(x) is false

  10. Example • If the square of an integer is even, then the integer is even. • Prove that for all integers n, if n2 is even then n is even. • Contrapositive: ∀integers n, n is not even then n2 is not even. • Proof: • Suppose n is any odd integer. n = 2k + 1 • n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 • n2 = 2r + 1 (by definition n2 is odd) • hence, the contrapositive is true therefore the statement must be true—”if n is even then n2 is even”

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